Pips answers for Monday, December 8, 2025

Start with the “=” region at the top-left: all 3 cells must show the same number, and only 4s appear on enough dominoes to fill it.

• the “=” region at the top-left: 4, 4, 4
• the single-cell “<1” region at the left: 0
• the single-cell 2 region at the right: 2

1. The “=” region at the top-left needs every pip to match — nothing else fits. Place the 4–4 domino horizontally on row 1, column 1 and row 1, column 2.
2. The single-cell “<1” region at the left must total less than 1 — nothing else fits. Place the 0–3 domino vertically on row 2, column 1 and row 3, column 1.
3. The single-cell 2 region at the right needs a total of exactly 2 — nothing else fits. Place the 2–4 domino vertically on row 1, column 4 and row 2, column 4.
4. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 3; the “=” region at the top-left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 4–6 domino horizontally on row 2, column 2 and row 2, column 3.

Start with the “=” region at the top-right: all 3 cells must show the same number, and only 1 or 2 or 4 or 6 appear on enough dominoes to fill it.

• the single-cell “>4” region at the top: 6
• the “=” region at the top-right: 2, 2, 2
• the “=” region at the center: 1, 1, 1
• the single-cell “>4” region at the left: 6
• the single-cell 3 region at the center: 3
• the “=” region at the bottom-right: 4, 4
• the single-cell “>1” region at the bottom-left: 6

1. The “=” region at the top-right needs every pip to match — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–6 domino horizontally on row 1, column 3 and row 1, column 4.
2. The “=” region at the top-right needs every pip to match (1 of its 3 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–2 domino horizontally on row 2, column 2 and row 2, column 3.
3. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the “=” region at the top-right needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 2–4 domino vertically on row 2, column 4 and row 3, column 4.
4. The “=” region at the center needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 1–3 domino horizontally on row 3, column 2 and row 3, column 3.
5. The single-cell “>1” region at the bottom-left must total more than 1 — nothing else fits. Place the 1–6 domino horizontally on row 4, column 1 and row 4, column 2.
6. Row 3, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the single-cell “>4” region at the left must total more than 4 — nothing else fits. Place the 4–6 domino vertically on row 2, column 1 and row 3, column 1.
7. Row 4, column 4 has only one open neighbor, so its domino must run to row 4, column 3; the “=” region at the bottom-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–4 domino horizontally on row 4, column 3 and row 4, column 4.

This board has more than one valid layout; the steps follow the official answer.

Start with the 0 region at the right: the only way to make it with these dominoes is 0 + 0 + 0.

• the 21 region at the top-left: 5, 5, 5, 6
• the 7 region at the left: 3, 4
• the single-cell “>2” region at the center: 3
• the 0 region at the right: 0, 0, 0
• the “=” region at the bottom-right: 4, 4
• the “=” region at the bottom-left: 1, 1
• the 12 region at the bottom: 6, 6
• the “=” region at the bottom-right: 2, 2, 2, 2
• the “=” region at the bottom-left: 0, 0

1. The 0 region at the right needs a total of exactly 0 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–0 domino horizontally on row 3, column 6 and row 3, column 7.
2. The “=” region at the bottom-right needs every pip to match — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–2 domino horizontally on row 6, column 6 and row 6, column 7.
3. The 12 region at the bottom needs a total of exactly 12 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino vertically on row 4, column 4 and row 5, column 4.
4. The “=” region at the bottom-left needs every pip to match — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–6 domino horizontally on row 6, column 3 and row 6, column 4.
5. The 21 region at the top-left needs a total of exactly 21 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–5 domino horizontally on row 1, column 3 and row 1, column 4.
6. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 21 region at the top-left needs a total of exactly 21 (2 of its 4 cells are already filled) — nothing else fits. Place the 3–6 domino vertically on row 2, column 4 and row 3, column 4.
7. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 1; the 21 region at the top-left needs a total of exactly 21 (3 of its 4 cells are already filled) — nothing else fits. Place the 2–5 domino horizontally on row 1, column 1 and row 1, column 2.
8. Row 5, column 6 has only one open neighbor, so its domino must run to row 4, column 6; the “=” region at the bottom-right needs every pip to match (2 of its 4 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–2 domino vertically on row 4, column 6 and row 5, column 6.
9. The “=” region at the bottom-right needs every pip to match — nothing else fits. Place the 2–4 domino vertically on row 5, column 8 and row 6, column 8.
10. Row 3, column 8 has only one open neighbor, so its domino must run to row 4, column 8; the 0 region at the right needs a total of exactly 0 (2 of its 3 cells are already filled) — nothing else fits. Place the 0–4 domino vertically on row 3, column 8 and row 4, column 8.
11. The “=” region at the bottom-left needs every pip to match — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–1 domino horizontally on row 6, column 1 and row 6, column 2.
12. The 7 region at the left needs a total of exactly 7 — nothing else fits. Place the 1–4 domino vertically on row 4, column 1 and row 5, column 1.
13. Row 3, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the 7 region at the left needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–3 domino vertically on row 2, column 1 and row 3, column 1.